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SSC Pure Mathematics QUESTION #11161
Question 1
$\sec\!\left(\tan^{-1}\dfrac{2}{3}\right) =$
  • $\dfrac{2}{\sqrt{13}}$
  • $\dfrac{3}{\sqrt{13}}$
  • $\dfrac{\sqrt{13}}{3}$✔️
  • $\dfrac{\sqrt{13}}{2}$
Correct Answer Explanation
Let $\theta = \tan^{-1}(2/3)$, so $\tan\theta = 2/3$. Draw a right triangle: opposite $= 2$, adjacent $= 3$, hypotenuse $= \sqrt{4+9} = \sqrt{13}$. Therefore $\sec\theta = \dfrac{\text{hyp}}{\text{adj}} = \dfrac{\sqrt{13}}{3}$.